Question

The value of ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{3}$
• E. $0$

Answer 1

The correct option is C $\frac{\pi }{2}$

We need to find value of ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)$

Let ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)=t$

$\Rightarrow \sin t=\frac{2\sqrt{2}}{3}{\sin }^{2}t+{\cos }^{2}t=1{\left(\frac{2\sqrt{2}}{3}\right)}^2+{\cos }^{2}t=1\Rightarrow {\cos }^{2}t=\frac{1}{9}\Rightarrow \cos t=\frac{1}{3}\Rightarrow t={\cos }^{-1}\left(\frac{1}{3}\right)$

Substituting $t$

$\Rightarrow {\cos }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{2}$

So, option C is correct.